এ অধ্যায়ের পাঠ্যসূচী
- সংযোজিত ফাংশনের অন্তরীকরণ (Differentiation of composite functions)
- অনুসিদ্ধান্ত (Illustration)
- বিপরীত বৃত্তীয় ফাংশনের অন্তরীকরণ (The differentiation of inverse circular function)
- বিপরীত বৃত্তীয় ফাংশনের সূত্রসমুহ (Formulas for inverse circular functions)
- সংযোজিত এবং বিপরীত বৃত্তীয় ফাংশনের অন্তরজ নির্ণয়ের কৌশল (Techniques for Differentiating of Additive and Inverse Circular Functions)
- কোণ পরিমাপের তিনটি পদ্ধতি (Three methods of angle measurement)
- অধ্যায় \(ix.D\)-এর উদাহরণসমুহ
- অধ্যায় \(ix.D\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
- অধ্যায় \(ix.D\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(ix.D\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(ix.D\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
সংযোজিত ফাংশনের অন্তরীকরণ
Differentiation of composite functions
যখন, \(y=f(z)\) এবং \(z=g(x)\).
\(\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}\)
\(\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\)
অনুসিদ্ধান্ত
Illustration
যখন, \(y=f(z)\), \(z=g(t)\) এবং \(t=h(x)\).
\(\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dt}.\frac{dt}{dx}\)
যখন, \(y=f(z)\), \(z=g(t)\), \(t=h(u)\) এবং \(u=\psi(x)\).
\(\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dt}.\frac{dt}{du}.\frac{du}{dx}\)
বিপরীত বৃত্তীয় ফাংশনের অন্তরীকরণ
The differentiation of inverse circular function
\(\frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}\)
\(\frac{d}{dx}(\cos^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\)
\(\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\)
\(\frac{d}{dx}(\cos^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\)
\(\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\)
\(\frac{d}{dx}(cosec^{-1} \ x)=-\frac{1}{x\sqrt{x^2-1}}\)
\(\frac{d}{dx}(\sec^{-1} x)=\frac{1}{x\sqrt{x^2-1}}\)
\(\frac{d}{dx}(\cot^{-1} x)=-\frac{1}{1+x^2}\)
\(\frac{d}{dx}(\sec^{-1} x)=\frac{1}{x\sqrt{x^2-1}}\)
\(\frac{d}{dx}(\cot^{-1} x)=-\frac{1}{1+x^2}\)
বিপরীত বৃত্তীয় ফাংশনের সূত্রসমুহ
Formulas for inverse circular functions
\(\sin(\sin^{-1} x)=x\)
\(\sin^{-1}(\sin x)=x\)
\(\cos(\cos^{-1} x)=x\)
\(\cos^{-1}(\cos x)=x\)
\(\tan(\tan^{-1} x)=x\)
\(\tan^{-1}(\tan x)=x\)
\(\sin^{-1} x+\cos^{-1} x=\frac{\pi}{2}\)
\(\tan^{-1} x+\cot^{-1} x=\frac{\pi}{2}\)
\(\sin^{-1}(\sin x)=x\)
\(\cos(\cos^{-1} x)=x\)
\(\cos^{-1}(\cos x)=x\)
\(\tan(\tan^{-1} x)=x\)
\(\tan^{-1}(\tan x)=x\)
\(\sin^{-1} x+\cos^{-1} x=\frac{\pi}{2}\)
\(\tan^{-1} x+\cot^{-1} x=\frac{\pi}{2}\)
\(\sec^{-1} x+ cosec^{-1} \ x=\frac{\pi}{2}\)
\(\tan^{-1} x+\tan^{-1} y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)\)
\(\tan^{-1} x-\tan^{-1} y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)\)
\(2\tan^{-1} x=\sin^{-1}\left(\frac{2x}{1+x^2}\right)\)
\(2\tan^{-1} x=\tan^{-1}\left(\frac{2x}{1-x^2}\right)\)
\(2\tan^{-1} x=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
\(\tan^{-1} x+\tan^{-1} y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)\)
\(\tan^{-1} x-\tan^{-1} y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)\)
\(2\tan^{-1} x=\sin^{-1}\left(\frac{2x}{1+x^2}\right)\)
\(2\tan^{-1} x=\tan^{-1}\left(\frac{2x}{1-x^2}\right)\)
\(2\tan^{-1} x=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)
সংযোজিত এবং বিপরীত বৃত্তীয় ফাংশনের অন্তরজ নির্ণয়ের কৌশল
Techniques for Differentiating of Additive and Inverse Circular Functions
\(f(x)\) বীজগানিতিক ফাংশনের অন্তরজ নির্ণয়ের ক্ষেত্রে, বিপরীত বৃত্তীয় ফাংশনে রূপান্তর করে সরল করণের পদ্ধতিঃ
\(f(x)\)-এর আকার
\(\sqrt{a^2-x^2}\)
\(\sqrt{1-x^2}\)
\(\sqrt{a^2+x^2}\)
\(\sqrt{1+x^2}\)
\(\sqrt{x^2-a^2}\)
\(\sqrt{x^2-1}\)
\(\sqrt{a+x}\) এবং \(\sqrt{a-x}\)
\(\sqrt{a^2+x^2}\) এবং \(\sqrt{a^2-x^2}\)
\(\frac{2x}{1-x^2}\) অথবা \(\frac{2x}{1+x^2}\) অথবা \(\frac{1-x^2}{1+x^2}\)
\(\frac{1+x}{1-x}\) অথবা \(\frac{1-x}{1+x}\)
\(\sqrt{\frac{1+x}{1-x}}\) অথবা \(\sqrt{\frac{1-x}{1+x}}\)
\(\sqrt{a^2-x^2}\)
\(\sqrt{1-x^2}\)
\(\sqrt{a^2+x^2}\)
\(\sqrt{1+x^2}\)
\(\sqrt{x^2-a^2}\)
\(\sqrt{x^2-1}\)
\(\sqrt{a+x}\) এবং \(\sqrt{a-x}\)
\(\sqrt{a^2+x^2}\) এবং \(\sqrt{a^2-x^2}\)
\(\frac{2x}{1-x^2}\) অথবা \(\frac{2x}{1+x^2}\) অথবা \(\frac{1-x^2}{1+x^2}\)
\(\frac{1+x}{1-x}\) অথবা \(\frac{1-x}{1+x}\)
\(\sqrt{\frac{1+x}{1-x}}\) অথবা \(\sqrt{\frac{1-x}{1+x}}\)
প্রতিস্থাপন
\(x=a\sin \theta\) অথবা \(x=a\cos \theta\)
\(x=\sin \theta\) অথবা \(x=\cos \theta\)
\(x=a\tan \theta\) অথবা \(x=a\cot \theta\)
\(x=\tan \theta\) অথবা \(x=\cot \theta\)
\(x=a\sec \theta\) অথবা \(x=a \ cosec \ \theta\)
\(x=\sec \theta\) অথবা \(x= cosec \ \theta\)
\(x=a\cos 2\theta\)
\(x^2=a^2\cos 2\theta\)
\(x=\tan \theta\)
\(x=\tan \theta\)
\(x=\cos \theta\)
\(x=a\sin \theta\) অথবা \(x=a\cos \theta\)
\(x=\sin \theta\) অথবা \(x=\cos \theta\)
\(x=a\tan \theta\) অথবা \(x=a\cot \theta\)
\(x=\tan \theta\) অথবা \(x=\cot \theta\)
\(x=a\sec \theta\) অথবা \(x=a \ cosec \ \theta\)
\(x=\sec \theta\) অথবা \(x= cosec \ \theta\)
\(x=a\cos 2\theta\)
\(x^2=a^2\cos 2\theta\)
\(x=\tan \theta\)
\(x=\tan \theta\)
\(x=\cos \theta\)
কোণ পরিমাপের তিনটি পদ্ধতি
Three methods of angle measurement
এদের মধ্যে সম্পর্কঃ
×
যখন, \(y=f(z)\) এবং \(z=g(x)\).
\((1)\) \(\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}\)
proof:
মনে করি,
\(x\)-এর অতিক্ষুদ্র বৃদ্ধি \(\delta x\)-এর জন্য, \(y\) ও \(z\) -এ বৃদ্ধি যথাক্রমে \(\delta y\) এবং \(\delta z\);
যেহেতু এই বৃদ্ধিগুলি সসীম,
অতএব,
\(\frac{\delta y}{\delta x}=\frac{\delta y}{\delta z}.\frac{\delta z}{\delta x}\)
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x\rightarrow 0}\left(\frac{\delta y}{\delta z}.\frac{\delta z}{\delta x}\right)\]
\[=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta z}.\lim_{\delta x\rightarrow 0}\frac{\delta z}{\delta x}\]
যেহেতু \(z, x\)-এর একটি ফাংশন,
অতএব,
\(\delta x\rightarrow 0\) হলে \(\delta z\rightarrow 0\) হবে।
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta z}.\lim_{\delta x\rightarrow 0}\frac{\delta z}{\delta x}\]
\[\Rightarrow \frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}\] ➜ \[\because \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\frac{dy}{dx}\]
(proved)
\(\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}\)
\(x\)-এর অতিক্ষুদ্র বৃদ্ধি \(\delta x\)-এর জন্য, \(y\) ও \(z\) -এ বৃদ্ধি যথাক্রমে \(\delta y\) এবং \(\delta z\);
যেহেতু এই বৃদ্ধিগুলি সসীম,
অতএব,
\(\frac{\delta y}{\delta x}=\frac{\delta y}{\delta z}.\frac{\delta z}{\delta x}\)
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x\rightarrow 0}\left(\frac{\delta y}{\delta z}.\frac{\delta z}{\delta x}\right)\]
\[=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta z}.\lim_{\delta x\rightarrow 0}\frac{\delta z}{\delta x}\]
যেহেতু \(z, x\)-এর একটি ফাংশন,
অতএব,
\(\delta x\rightarrow 0\) হলে \(\delta z\rightarrow 0\) হবে।
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta z}.\lim_{\delta x\rightarrow 0}\frac{\delta z}{\delta x}\]
\[\Rightarrow \frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}\] ➜ \[\because \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\frac{dy}{dx}\]
(proved)
×
যখন, \(y=f(z)\) এবং \(z=g(x)\).
\((2)\) \(\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\)
proof:
মনে করি,
\(x\)-এর অতিক্ষুদ্র বৃদ্ধি \(\delta x\)-এর জন্য, \(y\)-এর বৃদ্ধি \(\delta y\);
যেহেতু এই বৃদ্ধিগুলি সসীম,
অতএব,
\(\frac{\delta y}{\delta x}=\frac{1}{\frac{\delta x}{\delta y}}\)
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}= \lim_{\delta x\rightarrow 0}\frac{1}{\frac{\delta x}{\delta y}}\]
\[\Rightarrow \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}= \frac{1}{\lim_{\delta x\rightarrow 0}\frac{\delta x}{\delta y}}\]
\[\Rightarrow \frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}\] ➜ \[\because \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\frac{dy}{dx}\]
(proved)
\(\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\)
\(x\)-এর অতিক্ষুদ্র বৃদ্ধি \(\delta x\)-এর জন্য, \(y\)-এর বৃদ্ধি \(\delta y\);
যেহেতু এই বৃদ্ধিগুলি সসীম,
অতএব,
\(\frac{\delta y}{\delta x}=\frac{1}{\frac{\delta x}{\delta y}}\)
\[\therefore \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}= \lim_{\delta x\rightarrow 0}\frac{1}{\frac{\delta x}{\delta y}}\]
\[\Rightarrow \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}= \frac{1}{\lim_{\delta x\rightarrow 0}\frac{\delta x}{\delta y}}\]
\[\Rightarrow \frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}\] ➜ \[\because \lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\frac{dy}{dx}\]
(proved)
×
\((1)\) \(\frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}\)
proof:
মনে করি,
\(\sin^{-1} x=y\)
\(\Rightarrow x=\sin y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\sin y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\cos y\)
\(=\sqrt{1-\sin^2 y}\) ➜ \(\because \cos A=\sqrt{1-\sin^2 A}\)
\(=\sqrt{1-x^2}\) ➜ \(\because x=\sin y\)
\(\therefore \frac{dx}{dy}=\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}\)
(proved)
\(\frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}\)
\(\sin^{-1} x=y\)
\(\Rightarrow x=\sin y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\sin y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\cos y\)
\(=\sqrt{1-\sin^2 y}\) ➜ \(\because \cos A=\sqrt{1-\sin^2 A}\)
\(=\sqrt{1-x^2}\) ➜ \(\because x=\sin y\)
\(\therefore \frac{dx}{dy}=\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}\)
(proved)
×
\((2)\) \(\frac{d}{dx}(\cos^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\)
proof:
মনে করি,
\(\cos^{-1} x=y\)
\(\Rightarrow x=\cos y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\cos y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-\sin y\)
\(=-\sqrt{1-\cos^2 y}\) ➜ \(\because \sin A=\sqrt{1-\cos^2 A}\)
\(=-\sqrt{1-x^2}\) ➜ \(\because x=\cos y\)
\(\therefore \frac{dx}{dy}=-\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{d}{dx}(\cos^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\)
(proved)
\(\frac{d}{dx}(\cos^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\)
\(\cos^{-1} x=y\)
\(\Rightarrow x=\cos y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\cos y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-\sin y\)
\(=-\sqrt{1-\cos^2 y}\) ➜ \(\because \sin A=\sqrt{1-\cos^2 A}\)
\(=-\sqrt{1-x^2}\) ➜ \(\because x=\cos y\)
\(\therefore \frac{dx}{dy}=-\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{d}{dx}(\cos^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\)
(proved)
×
\((3)\) \(\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\)
proof:
মনে করি,
\(\tan^{-1} x=y\)
\(\Rightarrow x=\tan y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\tan y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\sec^2 y\)
\(=1+\tan^2 y\) ➜ \(\because \sec^2 A=1+\tan^2 A\)
\(=1+x^2\) ➜ \(\because x=\tan y\)
\(\therefore \frac{dx}{dy}=1+x^2\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{1+x^2}\)
\(\therefore \frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\)
(proved)
\(\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\)
\(\tan^{-1} x=y\)
\(\Rightarrow x=\tan y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\tan y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\sec^2 y\)
\(=1+\tan^2 y\) ➜ \(\because \sec^2 A=1+\tan^2 A\)
\(=1+x^2\) ➜ \(\because x=\tan y\)
\(\therefore \frac{dx}{dy}=1+x^2\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{1+x^2}\)
\(\therefore \frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}\)
(proved)
×
\((4)\) \(\frac{d}{dx}(cosec^{-1} \ x)=-\frac{1}{x\sqrt{x^2-1}}\)
proof:
মনে করি,
\(cosec^{-1} \ x=y\)
\(\Rightarrow x=cosec \ y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(cosec \ y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-cosec \ y\cot y\)
\(=-cosec \ y\sqrt{cosec^2 \ y-1}\) ➜ \(\because \cot A=\sqrt{cosec^2 \ A-1}\)
\(=-x\sqrt{x^2-1}\) ➜ \(\because x=cosec \ y\)
\(\therefore \frac{dx}{dy}=-x\sqrt{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{x\sqrt{x^2-1}}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{d}{dx}(cosec^{-1} \ x)=-\frac{1}{x\sqrt{x^2-1}}\)
(proved)
\(\frac{d}{dx}(cosec^{-1} \ x)=-\frac{1}{x\sqrt{x^2-1}}\)
\(cosec^{-1} \ x=y\)
\(\Rightarrow x=cosec \ y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(cosec \ y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-cosec \ y\cot y\)
\(=-cosec \ y\sqrt{cosec^2 \ y-1}\) ➜ \(\because \cot A=\sqrt{cosec^2 \ A-1}\)
\(=-x\sqrt{x^2-1}\) ➜ \(\because x=cosec \ y\)
\(\therefore \frac{dx}{dy}=-x\sqrt{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{x\sqrt{x^2-1}}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{d}{dx}(cosec^{-1} \ x)=-\frac{1}{x\sqrt{x^2-1}}\)
(proved)
×
\((5)\) \(\frac{d}{dx}(\sec^{-1} x)=\frac{1}{x\sqrt{x^2-1}}\)
proof:
মনে করি,
\(\sec^{-1} x=y\)
\(\Rightarrow x=\sec y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\sec y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\sec y\tan y\)
\(=\sec y\sqrt{\sec^2 y-1}\) ➜ \(\because \tan A=\sqrt{\sec^2 A-1}\)
\(=x\sqrt{x^2-1}\) ➜ \(\because x=\sec y\)
\(\therefore \frac{dx}{dy}=x\sqrt{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{d}{dx}(\sec^{-1} x)=\frac{1}{x\sqrt{x^2-1}}\)
(proved)
\(\frac{d}{dx}(\sec^{-1} x)=\frac{1}{x\sqrt{x^2-1}}\)
\(\sec^{-1} x=y\)
\(\Rightarrow x=\sec y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\sec y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\sec y\tan y\)
\(=\sec y\sqrt{\sec^2 y-1}\) ➜ \(\because \tan A=\sqrt{\sec^2 A-1}\)
\(=x\sqrt{x^2-1}\) ➜ \(\because x=\sec y\)
\(\therefore \frac{dx}{dy}=x\sqrt{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{d}{dx}(\sec^{-1} x)=\frac{1}{x\sqrt{x^2-1}}\)
(proved)
×
\((6)\) \(\frac{d}{dx}(\cot^{-1} x)=-\frac{1}{1+x^2}\)
proof:
মনে করি,
\(\cot^{-1} x=y\)
\(\Rightarrow x=\cot y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\cot y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-cosec^2 \ y\)
\(=-(1+\cot^2 y)\) ➜ \(\because cosec^2 \ A=1+\cot^2 A\)
\(=-(1+x^2)\) ➜ \(\because x=\cot y\)
\(\therefore \frac{dx}{dy}=-(1+x^2)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{1+x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{1+x^2}\)
\(\therefore \frac{d}{dx}(\cot^{-1} x)=-\frac{1}{1+x^2}\)
(proved)
\(\frac{d}{dx}(\cot^{-1} x)=-\frac{1}{1+x^2}\)
\(\cot^{-1} x=y\)
\(\Rightarrow x=\cot y\)
\(\therefore \frac{d}{dy}(x)=\frac{d}{dy}(\cot y)\) ➜ \(y\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=-cosec^2 \ y\)
\(=-(1+\cot^2 y)\) ➜ \(\because cosec^2 \ A=1+\cot^2 A\)
\(=-(1+x^2)\) ➜ \(\because x=\cot y\)
\(\therefore \frac{dx}{dy}=-(1+x^2)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{1+x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{1}{1+x^2}\)
\(\therefore \frac{d}{dx}(\cot^{-1} x)=-\frac{1}{1+x^2}\)
(proved)
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